For the reaction
2 A(g) + B(g) → 2D(g)
= –10.5 kJ and 
= –44.1 JK–1.
Calculate 
for the reaction, and predict whether the reaction may occur spontaneously.
∆U = -10.5 kJ
Also, ∆U = ∆H – ∆nRT
Where,
∆U = change in energy
∆H = change in enthalpy
∆n = moles of products – moles of reactants = 2-3 = -1
R = 8.314 × 10-3 kJ mol-1 K-1 - constant
T = temperature; here room temperature is taken as 298 K
Substituting all the above values, we get
∆H = -10.5kJ + (-1 mol × (8.314× 10-3kJ mol-1 K-1) × 298 K
= -12.977kJ
we know that
∆G = ∆H - T∆S, where
∆G = change in Gibbs energy
∆H = change in enthalpy, got as -12.977 kJ
∆S = change in entropy, given as -44.1 JK-1 = -44.1 × 10-3 kJK-1
T = temperature at which reaction occurs = 298K
Substituting the above values
∆G = -12.977 – (298 × -44.1 × 10-3)
= -12.977 – ( -13.14)
= 0.163kJ
As ∆G > 0, reaction doesn’t occur spontaneously. Reactions are spontaneous only when ∆G < 0.